Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to determine what the variance of rolling $5$ pairs of two dice are when the sums of all $5$ pairs are added up (i.e. ranging from $10$ to $60$).

My first question is, when I calculate the variance using $E[X^2]-E[X]^2$ I get $2.91$, but my Excel spreadsheet and other sites I've googled give $3.5$ with no explanation of what me taking place. Which one is correct?

Second, to calculate the variance of a random variable representing the sum of the $5$ pairs (i.e. between $10$ and $60$), is it simply $5 \times Var(X)$? What about the standard deviation, is it $\sigma \sqrt{n}$?

Last, is there any difference between calculating the dice sums as "$5$ pairs of $2$ dice" and "$10$ dice"? Will it make a practical difference? (I find it easier to calculate it as $10$ dice).

share|improve this question
1  
It is not the variance, but the expected value of a dice roll, $E(X)$ that is 3.5. –  wnvl Nov 14 '12 at 1:53
1  
The assumptions in the second and third part of the question are all correct. –  wnvl Nov 14 '12 at 1:55
    
So, in other words the standard deviation of 5 pairs of 2 dice and the standard deviation of 10 dice is 5.3759? Can you confirm? –  Imray Nov 14 '12 at 2:02

1 Answer 1

So we are tossing $10$ dice. Let $X_i$ be the result of the $i$-th toss. Let $Y=X_1+X_2+\cdots +X_{10}$. It seems that you want the variance of $Y$.

The variance of a sum of independent random variables is the sum of the variances. Now calculate the variance of $X_i$. This as usual is $E(X_i^2)-(E(X_i))^2$.

We know that $E(X_i)=3.5$. For $E(X_i^2)$, note that this is $$\frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2).$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.